What is the general solution for tan(θ)-1-sqrt(2)=sec(θ) ?

The general solution for tan(θ)-1-sqrt(2)=sec(θ) is θ= pi/4+pin

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tan(θ)-1-sqrt(2)=sec(θ)

Solution

tan (θ) - 1 - 2 = sec (θ)

Solution

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

tan (θ) - 1 - 2 = sec (θ)

Square both sides

(tan (θ) - 1 - 2)^{2} = sec^{2} (θ)

Subtract

sec^{2} (θ)

from both sides

(tan (θ) - 1 - 2)^{2} - sec^{2} (θ) = 0

Rewrite using trig identities

(- 1 - 2 + tan (θ))^{2} - sec^{2} (θ)

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

= (- 1 - 2 + tan (θ))^{2} - (tan^{2} (θ) + 1)

Simplify

(- 1 - 2 + tan (θ))^{2} - (tan^{2} (θ) + 1) : - 2 tan (θ) - 22 tan (θ) + 2 + 22

(- 1 - 2 + tan (θ))^{2} - (tan^{2} (θ) + 1)

(- 1 - 2 + tan (θ))^{2} = (- 1 - 2 + tan (θ)) (- 1 - 2 + tan (θ))

= (tan (θ) - 1 - 2) (tan (θ) - 1 - 2) - (tan^{2} (θ) + 1)

Expand

(- 1 - 2 + tan (θ)) (- 1 - 2 + tan (θ)) : tan^{2} (θ) - 2 tan (θ) - 22 tan (θ) + 3 + 22

(- 1 - 2 + tan (θ)) (- 1 - 2 + tan (θ))

Distribute parentheses

= (- 1) (- 1) + (- 1) (- 2) + (- 1) tan (θ) + (- 2) (- 1) + (- 2) (- 2) + (- 2) tan (θ) + tan (θ) (- 1) + tan (θ) (- 2) + tan (θ) tan (θ)

Apply minus-plus rules

(- a) (- b) = ab, + (- a) = - a

= 1 \cdot 1 + 1 \cdot 2 - 1 \cdot tan (θ) + 1 \cdot 2 + 22 - 2 tan (θ) - 1 \cdot tan (θ) - 2 tan (θ) + tan (θ) tan (θ)

Simplify

1 \cdot 1 + 1 \cdot 2 - 1 \cdot tan (θ) + 1 \cdot 2 + 22 - 2 tan (θ) - 1 \cdot tan (θ) - 2 tan (θ) + tan (θ) tan (θ) : tan^{2} (θ) - 2 tan (θ) - 22 tan (θ) + 3 + 22

1 \cdot 1 + 1 \cdot 2 - 1 \cdot tan (θ) + 1 \cdot 2 + 22 - 2 tan (θ) - 1 \cdot tan (θ) - 2 tan (θ) + tan (θ) tan (θ)

Group like terms

= - 1 \cdot tan (θ) - 2 tan (θ) - 1 \cdot tan (θ) - 2 tan (θ) + tan (θ) tan (θ) + 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 2 + 22

Add similar elements:

- 2 tan (θ) - 2 tan (θ) = - 22 tan (θ)

= - 1 \cdot tan (θ) - 22 tan (θ) - 1 \cdot tan (θ) + tan (θ) tan (θ) + 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 2 + 22

Add similar elements:

1 \cdot 2 + 1 \cdot 2 = 22

= - 1 \cdot tan (θ) - 22 tan (θ) - 1 \cdot tan (θ) + tan (θ) tan (θ) + 1 \cdot 1 + 22 + 22

Add similar elements:

- 1 \cdot tan (θ) - 1 \cdot tan (θ) = - 2 tan (θ)

= - 2 tan (θ) - 22 tan (θ) + tan (θ) tan (θ) + 1 \cdot 1 + 22 + 22

tan (θ) tan (θ) = tan^{2} (θ)

tan (θ) tan (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

tan (θ) tan (θ) = tan^{1 + 1} (θ)

= tan^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= tan^{2} (θ)

1 \cdot 1 = 1

1 \cdot 1

Multiply the numbers:

1 \cdot 1 = 1

= 1

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - 2 tan (θ) - 22 tan (θ) + tan^{2} (θ) + 1 + 22 + 2

Add the numbers:

1 + 2 = 3

= tan^{2} (θ) - 2 tan (θ) - 22 tan (θ) + 3 + 22

= tan^{2} (θ) - 2 tan (θ) - 22 tan (θ) + 3 + 22

= tan^{2} (θ) - 2 tan (θ) - 22 tan (θ) + 3 + 22 - (tan^{2} (θ) + 1)

- (tan^{2} (θ) + 1) : - tan^{2} (θ) - 1

- (tan^{2} (θ) + 1)

Distribute parentheses

= - (tan^{2} (θ)) - (1)

Apply minus-plus rules

+ (- a) = - a

= - tan^{2} (θ) - 1

= tan^{2} (θ) - 2 tan (θ) - 22 tan (θ) + 3 + 22 - tan^{2} (θ) - 1

Simplify

tan^{2} (θ) - 2 tan (θ) - 22 tan (θ) + 3 + 22 - tan^{2} (θ) - 1 : - 2 tan (θ) - 22 tan (θ) + 2 + 22

tan^{2} (θ) - 2 tan (θ) - 22 tan (θ) + 3 + 22 - tan^{2} (θ) - 1

Group like terms

= tan^{2} (θ) - 2 tan (θ) - 22 tan (θ) - tan^{2} (θ) + 3 + 22 - 1

Add similar elements:

tan^{2} (θ) - tan^{2} (θ) = 0

= - 2 tan (θ) - 22 tan (θ) + 3 + 22 - 1

Add/Subtract the numbers:

3 - 1 = 2

= - 2 tan (θ) - 22 tan (θ) + 2 + 22

= - 2 tan (θ) - 22 tan (θ) + 2 + 22

= - 2 tan (θ) - 22 tan (θ) + 2 + 22

2 + 22 - 2 tan (θ) - 22 tan (θ) = 0

Solve by substitution

2 + 22 - 2 tan (θ) - 22 tan (θ) = 0

Let:

tan (θ) = u

2 + 22 - 2 u - 22 u = 0

2 + 22 - 2 u - 22 u = 0 : u = 1

2 + 22 - 2 u - 22 u = 0

Move

2

to the right side

2 + 22 - 2 u - 22 u = 0

Subtract

2

from both sides

2 + 22 - 2 u - 22 u - 2 = 0 - 2

Simplify

22 - 2 u - 22 u = - 2

22 - 2 u - 22 u = - 2

Move

22

to the right side

22 - 2 u - 22 u = - 2

Subtract

22

from both sides

22 - 2 u - 22 u - 22 = - 2 - 22

Simplify

- 2 u - 22 u = - 2 - 22

- 2 u - 22 u = - 2 - 22

Factor

- 2 u - 22 u : - 2 (1 + 2) u

- 2 u - 22 u

Rewrite as

= - 1 \cdot 2 u - 2 u 2

Factor out common term

2 u

= - 2 u (1 + 2)

- 2 (1 + 2) u = - 2 - 22

Divide both sides by

- 2 (1 + 2)

- 2 (1 + 2) u = - 2 - 22

Divide both sides by

- 2 (1 + 2)

\frac{- 2 ( 1 + 2 ) u}{- 2 ( 1 + 2 )} = - \frac{2}{- 2 ( 1 + 2 )} - \frac{2 2}{- 2 ( 1 + 2 )}

Simplify

\frac{- 2 ( 1 + 2 ) u}{- 2 ( 1 + 2 )} = - \frac{2}{- 2 ( 1 + 2 )} - \frac{2 2}{- 2 ( 1 + 2 )}

Simplify

\frac{- 2 ( 1 + 2 ) u}{- 2 ( 1 + 2 )} : u

\frac{- 2 ( 1 + 2 ) u}{- 2 ( 1 + 2 )}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2 ( 1 + 2 ) u}{2 ( 1 + 2 )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{( 1 + 2 ) u}{( 1 + 2 )}

Cancel the common factor:

1 + 2

= u

Simplify

- \frac{2}{- 2 ( 1 + 2 )} - \frac{2 2}{- 2 ( 1 + 2 )} : 1

- \frac{2}{- 2 ( 1 + 2 )} - \frac{2 2}{- 2 ( 1 + 2 )}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 2 - 2 2}{- 2 ( 1 + 2 )}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

- 2 - 22 = - (2 + 22)

= \frac{2 + 2 2}{2 ( 1 + 2 )}

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

= \frac{2 ( 1 + 2 )}{2 ( 1 + 2 )}

Cancel

\frac{2 ( 1 + 2 )}{2 ( 1 + 2 )} : 1

\frac{2 ( 1 + 2 )}{2 ( 1 + 2 )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{1 + 2}{( 1 + 2 )}

Cancel the common factor:

1 + 2

= 1

= 1

u = 1

u = 1

u = 1

Substitute back

u = tan (θ)

tan (θ) = 1

tan (θ) = 1

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

General solutions for

tan (θ) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Combine all the solutions

θ = \frac{π}{4} + πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

tan (θ) - 1 - 2 = sec (θ)

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{4} + πn :

True

\frac{π}{4} + πn

Plug in

n = 1

\frac{π}{4} + π 1

For

tan (θ) - 1 - 2 = sec (θ)

plug in

θ = \frac{π}{4} + π 1

tan (\frac{π}{4} + π 1) - 1 - 2 = sec (\frac{π}{4} + π 1)

Refine

- 1.41421 \dots = - 1.41421 \dots

\Rightarrow True

θ = \frac{π}{4} + πn

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cos (2 θ) = \frac{20}{29}

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- 6 sin (3 x) = 0

sin(x)=(sqrt(2))/4

sin (x) = \frac{2}{4}

2tan(2x)-6=0

2 tan (2 x) - 6 = 0

Frequently Asked Questions (FAQ)

What is the general solution for tan(θ)-1-sqrt(2)=sec(θ) ?
The general solution for tan(θ)-1-sqrt(2)=sec(θ) is θ= pi/4+pin